Physics Word Problem: Arrow and moving target collision

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The discussion centers on solving a physics word problem involving the collision of an arrow and a moving target. Participants express confusion over the calculations presented, particularly regarding the conservation of momentum and the variables used. Key issues include the unclear meaning of certain velocities, specifically the 23 m/s figure, and the incorrect subtraction of acceleration from velocity in the calculations. Additionally, the need for proper LaTeX formatting to clarify the equations is emphasized. Overall, the conversation highlights the importance of accurate representation and understanding of physics equations in problem-solving.
phantom lancer
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Homework Statement



Find the 2 unknown variables.[/B]

Homework Equations



Am I doing this right? Can someone help me find the solution?[/B]

The Attempt at a Solution



To compute for the velocity of the target/arrow combination immediately after the collision:
M_bV_{bi}=M_bV_{bf}+M_wV_{wf}
M_bV_{bf}=M_bV_{bf}-M_w-M_wV_{wf}
V_{bf}=\frac M_bV_{bi}-M_wV_{wf} M_b
V_{af}=\frac M_aV_{ai}-M_tV_{tf} M_a
V_{af}=\frac M_aV_{ai}-M_i M_a

V_{af}=\frac (0.323)(23 m/s)+(0.52 kg) (0.323 kg)
V_{af)=26.6 m/s down

To compute for velocity of the target/arrow combination just before it strikes the ground.

v=6.5 m/s - (9.8 m/s^2)[/B]
 

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Can you explain a little better what you have done? The lack of proper LaTeX coding and the symbols that you use make your strategy hard to discern. Clearly you are attempting to conserve momentum through the collision, but I see no calculation of the target's speed just before the collision. What is the meaning of 23 m/s? It cannot be the speed of the target because it has to be less than 15 m/s and it cannot be the speed of the arrow because it is given as 25 m/s. Also, the last equation seems to subtract an acceleration from a velocity. That cannot be.
 
I inserted double hashes (#) to turn on the LaTeX engine, and some {} braces as necessary, but could not make sense of the last line.
The second line is also clearly wrong.
phantom lancer said:
##M_bV_{bi}=M_bV_{bf}+M_wV_{wf}##
##M_bV_{bf}=M_bV_{bf}-M_w-M_wV_{wf}##
##V_{bf}=\frac {M_bV_{bi}-M_wV_{wf}}{ M_b}##
##V_{af}=\frac {M_aV_{ai}-M_tV_{tf}} {M_a}##
V_{af}=\frac M_aV_{ai}-M_i M_a
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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